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Accordingly to chapter 10, section 10.6 Feynman Rules of 'Introduction to Elementary Particles' by David Griffiths, there is a way to extract the vertex and propagators just by inspection of the Lagrangian:

1) Propagators: take the Euler-Lagrange equations of the free fields and the inverse of the operators in momentum space that act on the fields and multiplied by $i$ are the propagators for each one.

2) Vertex: take interaction Lagrangian and multiply it by $i$. Make use of the prescription $i\partial_\mu \rightarrow k_\mu$ and rub out the fields. The remaining is the vertex.

My questions are:

a) If I want to compute the vertex for the interaction Lagrangian ${\cal L}_{int} = g\varphi\partial_\mu \varphi \partial^\mu\varphi$ (all scalar fields), by rule 2) I get ${\rm vertex} = -igk_1k_2$. Nevertheless, apparently the solution is $-2ig(k_1k_2 + k_1k_3 + k_2k_3)$. Where did these extra factors come out?

b) If I had a similar interaction as in a) but changing one field for a new one $\chi$ (scalar too), so ${\cal L}_{int} = g\chi \partial_\mu \varphi \partial^\mu\varphi$, would the solution be $-2igk_1k_2$ with $k_i$ the momenta of $\varphi$ fields?

c) This book grants you that for QCD, with ${\cal L}_{int}^{3\ fields} = g\{[\partial^\mu A^\nu - \partial^\nu A^\mu]·(A_\mu \times A_\nu) + (A^\mu \times A^\nu)·[\partial_\mu A_\nu - \partial_\nu A_\mu]\}$, you can obtain the 3 fields vertex known as ${\rm vertex} = -gf^{\alpha \beta \gamma}[g_{\mu \nu}(k_1 - k_2)_\lambda + g_{\nu \lambda}(k_2 - k_3)_\mu + g_{\lambda \mu}(k_3 - k_1)_\nu]$. I've tried to get it from rule 2) but I wasn't able to.

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    $\begingroup$ a) Notice that you arbitrarily chose $k_1 k_2$ while you just have well could have chosen $k_1 k_3$ or $k_2 k__1$, etc. There are six ways to choose two momenta out of three, so you end up with six terms. Then you notice that $k_i k_j = k_j k_i$ so you can condense it into three terms times 2. Srednicki’s book does a good job of explaining Feynman rules and is free online. $\endgroup$ – Diffycue Nov 8 '18 at 1:55
  • $\begingroup$ Yes, I know. I'm starting to work with it (case a) is one of the non-solved exercises of that book). Are the rules that I talk about explained in this book? I don't see it $\endgroup$ – Vicky Nov 8 '18 at 2:03
  • $\begingroup$ the Feynman rules are explained in the chapter on Feynman rules $\endgroup$ – Diffycue Nov 8 '18 at 18:34
  • $\begingroup$ I meant this tricks/rules for Feynman rules $\endgroup$ – Vicky Nov 8 '18 at 18:39
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I don't get how to apply these rules, when the interaction contains derivatives, either. If you are unsure you can go the long way. If you do not know how, here is how one could do it:

Try to evaluate (e.g. for the $g\phi \partial_\mu \phi \partial^{\mu} \phi$ interaction) the $\mathcal{O}(g)$ greens functions $G(x_1,x_2,x_3)$ of 3 external scalar particles, e.g. by using wicks theorem, and see what the vertex rule is. In position space you get 6 terms of the form: $-ig\int d^4x D(x_1-x) (\partial_\mu D)(x_2-x) (\partial^\mu D)(x_3-x)$. Then consider the momentum space greensfunction $\tilde{G}(k_1,k_2,k_3) = \int d^4x_1 \int d^4x_2 \int d^4x_3 G(x_1,x_2,x_3) e^{-ik_1x_1}e^{-ik_2x_2}e^{-ik_3x_3}$ (usual convention is all momenta ingoing, i.e. $e^{-ikx}$). This gives you the $g k_2 k_3$ term. When you do this for all the other contractions you get $3$ non equal contributions and $2$ each of the total $6$ are equal (therefore the factor $2$).

When you have different scalar fields it works almost the same, you just only consider contractions between same types of scalar fields.

The QCD case is also the same, however all $6$ are non equal, because they you have an additional space time and colour index. Therefore you get the $6$ terms.

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  • $\begingroup$ So, under your way, for b) case, I could make $D(x_1 - x) \rightarrow D^\chi(x_1 - x)$ (propagator for $\chi$) and therefore there will be just two possible contractions by switching $x_2$ and $x_3$, so my solution will be the right one, true? $\endgroup$ – Vicky Nov 8 '18 at 0:05
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    $\begingroup$ yes. You are correct. The two contractions are the same and you get $-2igk_2k_3$. $\endgroup$ – lomby Nov 8 '18 at 1:23

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